tutorial-set-4

Studocu is not sponsored or endorsed by any college or university Tutorial set-4 Elementary University Physics II (Carleton University) Studocu is not sponsored or endorsed by any college or university Tutorial set-4 Elementary University Physics II (Carleton University) Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

PHYS 1008 2017W Tutorial #4 All sections 1. A glass block with index of refraction n 1 = 2 . 00 is submerged in an unknown liquid with index of refraction n 2 . (a) A light ray inside the glass block is incident on the glass-liquid interface at an angle of θ = 40 . 0 ◦ as shown in figure (a). If the ray undergoes total internal reflection, what is the maximum value of n 2 ? (b) A second light ray inside the glass block is incident on the glass-liquid interface at an angle of θ = 35 . 0 ◦ as shown in figure (b). In this case, the transmission angle is 90 . 0 ◦ so that a ray is in the interface plane between the glass and the liquid. What is the critical angle and what is the index of refraction n 2 ? (c) A third light ray inside the glass block is incident on the glass-liquid interface at an angle of θ = 30 . 0 ◦ as shown in figure (c). What is the transmission angle θ ? Solution: (a) First note that the critical angle is necessarily less than or equal to 40 . 0 ◦ , that is θ c ≤ 40 . 0 ◦ . If we saturate this bound, we can find the maximum value for n 2 . n 2 max = n 1 sin 40 . 0 ◦ = (2 . 00) sin 40 . 0 ◦ = 1 . 29 Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

2. Two converging lenses are placed 60.0 cm apart. An object is placed 80.0 cm to the left of the first lens. The focal length of the first lens is f 1 = 20 . 0 cm and the focal length of the second lens is f 2 = 15 . 0 cm . (a) Where is the image from the first lens formed? (b) What is the magnification of the first lens? (c) Where is the image from the second lens formed? (d) What is the magnification of the second lens? (e) What is the total magnification? (f) Is the final image upright or inverted? (g) Draw a ray diagram of this situation. Solution: (a) Use the thin-lens equation for the first lens 1 p 1 + 1 q 1 = 1 f 1 q 1 = bracketleftbigg 1 f 1 - 1 p 1 bracketrightbigg − 1 q 1 = bracketleftbigg 1 20 . 0 - 1 80 . 0 bracketrightbigg − 1 q 1 = 26 . 7 cm i.e. 26.7 cm to the right of the first lens. Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

(b) The magnification of the first lens is m 1 = - q 1 p 1 = - 26 . 7 80 . 0 = - 0 . 333 (c) First we note that p 2 = 60 . 0 - q 1 = 60 . 0 - 26 . 7 = 33 . 3 cm. Then 1 p 2 + 1 q 2 = 1 f 2 q 2 = bracketleftbigg 1 f 2 - 1 p 2 bracketrightbigg − 1 q 2 = bracketleftbigg 1 15 . 0 - 1 33 . 3 bracketrightbigg − 1 q 2 = 27 . 3 cm i.e. 27.3 cm to the right of the second lens. (d) The magnification of the second lens is m 2 = - q 2 p 2 = - 27 . 3 33 . 3 = - 0 . 819 (e) The total magnification is m t = m 1 m 2 = ( - 0 . 333)( - 0 . 819) = 0 . 273 (f) Since the total magnification is positive, the final image is upright. (g) There are many different possible ray diagrams for this situation. Below we show 4 different examples. Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

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3. A person struggles to read a book that is too close to their eyes. They have to hold the book at arm’s length, a distance of 55 cm away from their eyes, in order to read it. (a) Is this person nearsighted or farsighted? (b) What power of reading glasses should be prescribed to this person? Assume that the glasses will be placed 2.0 cm away from their eyes and that the glasses allow the person to read at a normal near point of 25 cm. (c) Draw a ray diagram showing how the reading glass lens forms its image. Solution: (a) The person struggles to read a book that is close to their eyes which means that they are farsighted. (b) The person can see the book clearly if it is placed at their near point of 55 cm. When the person is wearing their glasses, the book must be placed at a normal near point of 25 cm (i.e . 23 cm away from the glasses). Thus the glasses must produce a virtual image 53 cm to the left of the glasses. It follows that P = 1 f g = 1 p + 1 q P = 1 0 . 23 + 1 - 0 . 53 P = 2 . 5 D Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

4. Two loudspeakers are separated by a distance D = 6 . 00 m and are in phase with each other. A young healthy listener sits at point A, which is directly in front of one speaker at a distance d 1 = 8 . 00 m, so that the two speakers and the listener form a right triangle. The speed of sound in this room is v s = 340 m/s. (a) For what lowest frequency will the sound waves from each speaker interfere destructively at point A? (b) Assume a healthy young person cannot hear frequencies higher than f max = 20 . 0 kHz. What is the highest frequency that will interfere con- structively at point A and that would still be hearable by the listener? Solution: (a) For destructive interference, we require the the path difference Δ l be- tween the two waves to be an half-integer multiple of the wavelength. The path differ- ence between the two waves is Δ l = d 2 - d 1 = parenleftBig radicalbig d 2 1 + D 2 parenrightBig - d 1 = 10 . 0 - 8 . 00 = 2 . 00 m. Since we want the lowest frequency, we set m = 0. Using the relationship between the speed, frequency, and wavelength, we get Δ l = parenleftbigg m + 1 2 parenrightbigg λ = parenleftbigg m + 1 2 parenrightbigg v s f f = parenleftbigg m + 1 2 parenrightbigg v s Δ l f = parenleftbigg 0 + 1 2 parenrightbigg (340) (2 . 00) f = 85 . 0 Hz (b) For constructive interference, we need the path difference to be an integer multiple of the wavelength. Let’s find what m should be to have constructive interference at Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

f = 20 . 0 kHz. Δ l = mλ = m v s f m = Δ lf v s m = (2 . 00)(20 . 0 × 10 3 ) (340) m = 117 . 6 But since m should be an integer, we round down to m = 117. Thus, the highest frequency that can be heard by the listener and that interferes constructively is Δ l = mλ = m v s f f = v s m Δ l f = (340)(117) (2 . 00) f = 19 . 9 kHz Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

(b) In this case we get y 2 = L parenleftbigg mλ 2 d parenrightbigg = (1 . 6) parenleftbigg (3)(520 × 10 − 9 ) 0 . 18 × 10 − 3 parenrightbigg = 14 mm Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633

6. Light falling perpendicularly on a 9700 line/cm diffraction grating is re- vealed to contain three maxima in the first-order spectrum. These three maxima are at angles θ 1 = 31 . 2 ◦ , θ 2 = 36 . 4 ◦ , and θ 3 = 41 . 3 ◦ respectively. (a) What is the distance between the slits in the grating? (b) What wavelengths do these maxima correspond to? Solution: (a) The distance between the slits in the grating is d = 1 9700 = 1 . 03 μ m (b) We find the wavelengths using d sin θ = mλ λ = d sin θ m = d sin θ λ 1 = (1 . 03 × 10 − 6 ) sin(31 . 2 ◦ ) = 534 nm λ 2 = (1 . 03 × 10 − 6 ) sin(36 . 4 ◦ ) = 612 nm λ 3 = (1 . 03 × 10 − 6 ) sin(41 . 3 ◦ ) = 680 nm Downloaded by Steven Dumba (stevensdumba@gmail.com) lOMoARcPSD|10162633